Let $\mathcal{M}$ be a symplectic manifold in $\mathbb{R}^4$ of codimension 2 with the symplectic 2-form $dx_1 \wedge dy_1 + dx_2 \wedge dy_2$. Suppose that $\mathcal{M}$ intersects the $(x_2,y_2)$-plane perpendicular in the origin. Does there exist a symplectic transformation $(x_2,y_2) \mapsto (\hat{x}_2,\hat{y}_2) $ such that for an open neighbourhood of the origin in $\mathcal{M}$ the $ \hat{x}_2,\hat{y}_2$ are constant zero?

When you specify perpendicular, do you mean with respect to the symplectic form? If so, I think the answer is yes and it's Lemma 2.3 from Gompf's timeless classic Annals paper "A New Construction of Symplectic Manifolds". Otherwise there is a Kaehler angle between two symplectic planes which is preserved by symplectomorphisms and which is precisely the obstruction to Gompf's construction. See, for example, one of the early sections of Donaldson's JDG "Symplectic submanifolds and almost complex geometry" paper for the definition of Kaehler angle.
–
Jonny EvansMay 4 '12 at 20:53

1 Answer
1

I am assuming you write perpendicular in body of the question to mean transversal as in its title. Now I read your question as searching for local functions $\hat{x}_2=\hat{x}_2(x_2,y_2),\hat{y}_2=\hat{y}_2(x_2,y_2)$ such that $dx_2\wedge dy_2=d\hat{x}_2\wedge d\hat{y}_2$ and $\mathcal{M}=\hat{x}_2^{-1}(0)\cap\hat{y}_2^{-1}(0)$ locally around $0$.

If this reading is correct then in general the answer is no.
Infact if there were local symplectic coordinates $(\hat{x}_2,\hat{y}_2)$ on the $(x_2,y_2)$ plane such that $\mathcal{M}\subseteq\hat{x}^{-1}_2(0)\cap\hat{y}^{-1}_2(0)$ in a neighborhood of $0$, then $\mathcal{M}$ should be locally included in the $(x_1,y_1)$ plane (because $\hat{x}^{-1}_2(0)\cap\hat{y}^{-1}_2(0)=x^{-1}_2(0)\cap y^{-1}_2(0)$.)

But one can construct a counter-example already taking symplectic vector subspaces complementary to the $(x_2,y_2)$-plane in the constant symplectic space $(\mathbb{R}^4,dx_1\wedge dy_1+dx_2\wedge dy_2)$.